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Q.
Let $A = \left\{0 \in \left( - \frac{\pi}{2} , \pi\right) : \frac{3+2 i \sin \theta}{1-2 i \sin\theta} \text{is}\, \text{purely} \,\text{imaginary}\right\}$ Then the sum of the elements in $A$ is :
JEE MainJEE Main 2019Complex Numbers and Quadratic Equations
Solution:
Given $z= \frac{3+2i\sin\theta}{1-2 i\sin\theta}$ is purely img
so real part becomes zero.
$ z = \left(\frac{3+2i \sin\theta}{1-2 i \sin\theta}\right) \times \left(\frac{1+2i \sin\theta}{1+2i\sin\theta}\right) $
$ z= \frac{\left(3-4 \sin^{2}\theta\right)+i\left(8\sin\theta\right)}{i+4\sin^{2} \theta } $
Now Re(z) = 0
$ \frac{3-4 \sin^{2} \theta}{1+4\sin^{2} \theta} = 0$
$ \sin^{2} \theta = \frac{3}{4} \sin\theta = \pm \frac{\sqrt{3}}{2} \Rightarrow \theta = - \frac{\pi}{3} , \frac{\pi}{3}, \frac{2\pi}{3} $
$ \because \theta \in\left( - \frac{\pi}{2} , \pi\right) $
then sum of the elements in A is
$ - \frac{\pi}{3} + \frac{\pi}{3} + \frac{2\pi}{3 } = \frac{2 \pi}{3} $